Answer
a. 100;
b. 1
Work Step by Step
We are given $S(t)=100-100t^{-1}$
a. For $t=1$, the rate of change is:
$\lim\limits_{h \to 0}\frac{s(1+h)-s(1)}{h}$
Finding $s(1+h)=100-100(1+h)^{-1}$
and $s(1)=100-100(1)^{-1}=0$
Therefore, the rate of change at $t=1$ is
$\lim\limits_{h \to 0}\frac{100-100(1+h)^{-1}}{h}=\lim\limits_{h \to 0} \frac{100-\frac{100}{1+h}}{h}=\lim\limits_{h \to 0}\frac{100}{1+h}=100$
b. For $t=10$, the rate of change is:
$\lim\limits_{h \to 0}\frac{s(10+h)-s(10)}{h}$
Finding $s(10+h)=100-100(10+h)^{-1}$
and $s(10)=100-100(10)^{-1}=90$
Therefore, the rate of change at $t=1$ is
$\lim\limits_{h \to 0}\frac{100-100(10+h)^{-1}-90}{h}=\lim\limits_{h \to 0} \frac{10-\frac{100}{10+h}}{h}=\lim\limits_{h \to 0}\frac{10}{10+h}=1$
